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**Quiz-3 / TUT5103 Quiz 3**

« **on:**October 11, 2019, 01:24:36 PM »

Q: Find the solution of the given initial value problem.

y'' + y' - 2y = 0, y(0) = 1, y'(0) = 1

Answer:

r^2+r-2 = 0

(r-1)(r-2) = 0

r = 1 or r = -2

y = C1e^t + C2e^(-2t)

hence, y'(t) = C1e^t -2C2e^(-2t)

Because y(0)= 1, C1+ C2 = 1

y'(0) = 1, C1-2C2 = 1

C1 = 1, C2 = 0

Thus, y = e^t

y'' + y' - 2y = 0, y(0) = 1, y'(0) = 1

Answer:

r^2+r-2 = 0

(r-1)(r-2) = 0

r = 1 or r = -2

y = C1e^t + C2e^(-2t)

hence, y'(t) = C1e^t -2C2e^(-2t)

Because y(0)= 1, C1+ C2 = 1

y'(0) = 1, C1-2C2 = 1

C1 = 1, C2 = 0

Thus, y = e^t